Boundary of boundary of open set is equal to the boundary of said set

Question

Let $A=A^{\circ}$. Proof that $$ \partial(\partial A)) = \partial A $$ By definition \begin{align} \partial(\partial A)) &=\overline{\partial A}\setminus (\partial A)^{\circ}&\textit{Since $\overline{\partial A}=\partial A$}\\ &=\partial A\setminus (\partial A)^{\circ}&\textit{By definition of $\partial$}\\ &=\partial A\setminus(\overline{A}\setminus A^{\circ})^{\circ}&\textit{Since $A=A^{\circ}$}\\ &=\partial A\setminus (\overline{A}\setminus A)^{\circ}&\textit{Take interior of both}\\ &=\partial A \setminus(A\setminus A)\\ &=\partial A\setminus\emptyset\\ &=\partial A \end{align} I am not quite sure that all the steps are legal. Can you help me out?

Answer 1

In general, it is not true that $(X\backslash Y)^{\circ} = X^\circ \backslash Y^\circ$. For example, take $X = \mathbb{R}$ and $Y = \{1\}$ under the usual topolgy in $\mathbb{R}$. So the step where you "take the interior of both" is not well justified. You also seem to have assumed that $(\bar{A})^\circ = A$. Again, this is not true.

An easier way is to prove what you want is by showing that the interior of a boundary of an open set is empty. That will directly give you that $\partial A \backslash \partial A^\circ = \partial A$.

(Big) Hint:

Use the fact that $p \in \partial A$ iff every neighborhood of $p$ contains points from $A$ and $A^C$. Since points of $A$ are not on the boundary, $p$ cannot be an interior point of $\partial A$.

Good luck!